Problem: $f(x) = 2x-4(g(x))$ $g(t) = 3t-1$ $ g(f(0)) = {?} $
First, let's solve for the value of the inner function, $f(0)$ . Then we'll know what to plug into the outer function. $f(0) = (2)(0)-4(g(0))$ To solve for the value of $f$ , we need to solve for the value of $g(0)$ $g(0) = (3)(0)-1$ $g(0) = -1$ That means $f(0) = (2)(0)+(-4)(-1)$ $f(0) = 4$ Now we know that $f(0) = 4$ . Let's solve for $g(f(0))$ , which is $g(4)$ $g(4) = (3)(4)-1$ $g(4) = 11$